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Question: The point of intersection of the curves whose parametric equations are \(x = {t^2} + 1\), \(y = 2t\)...

The point of intersection of the curves whose parametric equations are x=t2+1x = {t^2} + 1, y=2ty = 2t and x=2sx = 2s, y=2sy = \dfrac{2}{s} is given by
A.(4,1)\left( {4,1} \right)
B.(2,2)\left( {2,2} \right)
C.(−2,4)\left( { - 2,4} \right)
D.(1,2)\left( {1,2} \right)

Explanation

Solution

We will first find the value of tt from the given value of xx, then will substitute the value in the equation of yy involving tt. Similarly, find the value of ss from one equation and substitute in another to get an equation of xx and yy. Solve the two equations to determine the values of xx and yy.

Complete step-by-step answer:
First of all, we will find the value of tt from the given equation y=2ty = 2t
Then, t=y2t = \dfrac{y}{2}
Substitute the value of tt in the equation of xx, that is x=t2+1x = {t^2} + 1
Therefore, now we have x=(y2)2+1x = {\left( {\dfrac{y}{2}} \right)^2} + 1
On simplifying, we get,
x=y24+1x = \dfrac{{{y^2}}}{4} + 1
x=y2+44x = \dfrac{{{y^2} + 4}}{4} (1)
Similarly, we will find the value of ss from the equation y=2sy = \dfrac{2}{s}
Now, we will get s=2ys = \dfrac{2}{y}
Next, we will substitute the value of s=2ys = \dfrac{2}{y} in the equation x=2sx = 2s to find the value of xx.
Then, we have
x=2(2y)x = 2\left( {\dfrac{2}{y}} \right)
⇒x=4y\Rightarrow x = \dfrac{4}{y}
⇒xy=4\Rightarrow xy = 4 (2)
Now, we will equate equations (1) and (2) we will get,
(y2+44)y=4\left( {\dfrac{{{y^2} + 4}}{4}} \right)y = 4
Hence, on solving the equation we get,
y(y2+4)=16y\left( {{y^2} + 4} \right) = 16
Now, we add and subtract 4y4y in the bracket to complete the square and hence solve further,
y(y2+4−4y+4y)−16=0 y(y−2)2+4y2−16=0  y\left( {{y^2} + 4 - 4y + 4y} \right) - 16 = 0 \\\ y{\left( {y - 2} \right)^2} + 4{y^2} - 16 = 0 \\\
Now we will take terms common and simplify,
y(y−2)(y−2)+4(y2−4)=0 ⇒y(y−2)(y−2)+4(y+2)(y−2)=0 ⇒(y−2)(y2−2y+4y+8)=0 ⇒(y−2)(y2+2y+8)=0  y\left( {y - 2} \right)\left( {y - 2} \right) + 4\left( {{y^2} - 4} \right) = 0 \\\ \Rightarrow y\left( {y - 2} \right)\left( {y - 2} \right) + 4\left( {y + 2} \right)\left( {y - 2} \right) = 0 \\\ \Rightarrow \left( {y - 2} \right)\left( {{y^2} - 2y + 4y + 8} \right) = 0 \\\ \Rightarrow \left( {y - 2} \right)\left( {{y^2} + 2y + 8} \right) = 0 \\\
We will equate each of the factors to 0 and find the value of yy
That is, when y−2=0y - 2 = 0
We get, y=2y = 2
Similarly, when we put y2+2y+8=0{y^2} + 2y + 8 = 0
We will first calculate DD which is 4ac4ac
Hence, we have
D=b2−4ac ⇒D=(2)2−4(1)(9) ⇒D=4−36 ⇒D=−34  D = {b^2} - 4ac \\\ \Rightarrow D = {\left( 2 \right)^2} - 4\left( 1 \right)\left( 9 \right) \\\ \Rightarrow D = 4 - 36 \\\ \Rightarrow D = - 34 \\\
Since, D<0D < 0, there does not exist any real root corresponding to the factor y2+2y+8{y^2} + 2y + 8
Therefore, the value of yy is 2
Now, we will substitute the value of yy in equation (2) to find the value of xx
Thus,
x(2)=4 x=2  x\left( 2 \right) = 4 \\\ x = 2 \\\
Hence, the required point is (2,2)\left( {2,2} \right)
Thus, option B is correct.

Note: Parametric equations are a set of equations that represent the variables as a function of an independent variable. Since, there are two sets of equations involving two parameters, two equations will be formed. And the point of the intersection of the equations is the solution of the two equations.